📚 Year 7 SQA Maths Unit Test Mock Paper Analysis | Year 7 SQA 数学单元测试模拟卷解析
This article walks you through a full Year 7 SQA Maths unit test mock paper, covering number operations, fractions, decimals, percentages, algebra, angles, area, perimeter, coordinates and statistics. Every question is broken down with clear step-by-step reasoning, common mistake alerts and targeted revision advice. Whether you are preparing for an end‑of‑unit test or consolidating your S1 skills, this analysis will help you build confidence and accuracy.
本文带你完整解析一份 Year 7 SQA 数学单元测试模拟卷，覆盖整数运算、分数、小数、百分数、代数、角度、面积、周长、坐标和统计。每道题都配有清晰的逐步推理、常见错误提醒和针对性复习建议。无论你是在备战单元测验，还是巩固 S1 数学技能，这份解析都能帮你提升自信与准确率。
1. Mock Paper Overview | 模拟卷概览
This mock paper contains eight carefully chosen questions that mirror the style and difficulty of a typical SQA Year 7 unit test. The topics include: whole number calculations with BIDMAS, fraction addition and simplification, decimal‑percentage conversion and percentage of an amount, simplifying algebraic expressions and solving one‑step equations, angle properties in a triangle, area and perimeter of a rectangle, coordinate geometry and translation, and calculating the mean, median and mode from a small data set. Tackling this paper will give you a clear picture of your current strengths and the areas where extra practice is needed.
这份模拟卷包含八道精心挑选的题目，贴近 SQA Year 7 单元测验的风格和难度。主题涵盖：运用 BIDMAS 法则的整数运算，分数加法与约分，小数与百分数互化及求一个数的百分之几，简化代数表达式与解一步方程，三角形内角性质，长方形面积与周长，坐标几何与平移，以及根据小数据集计算平均数、中位数和众数。完成这份试卷能让你清楚地看到自己当前的强项和需要额外练习的领域。
2. Question 1: Order of Operations | 问题1：整数运算顺序
Calculate: 45 + 72 ÷ 8 − 15 × 2
To evaluate this expression correctly, we must follow the order of operations, often remembered by the acronym BIDMAS or BODMAS. The letters stand for Brackets, Indices, Division, Multiplication, Addition and Subtraction. Division and multiplication have equal priority and should be done before addition and subtraction, working from left to right.
要正确计算这个式子，必须遵循运算顺序，通常记作 BIDMAS 或 BODMAS。这几个字母分别代表括号、指数、除法、乘法、加法和减法。除法和乘法优先级相同，并且要在加法和减法之前运算，按照从左到右的顺序计算。
Step 1: Identify the division and multiplication. Here, 72 ÷ 8 and 15 × 2 are the operations with higher priority. Compute them: 72 ÷ 8 = 9, and 15 × 2 = 30.
步骤1：先找出除法和乘法。这里 72 ÷ 8 和 15 × 2 是优先级较高的运算。分别计算：72 ÷ 8 = 9，15 × 2 = 30。
Step 2: Substitute these results back into the expression. The original 45 + 72 ÷ 8 − 15 × 2 becomes 45 + 9 − 30.
步骤2：将这些结果代回原式。原来的 45 + 72 ÷ 8 − 15 × 2 变为 45 + 9 − 30。
Step 3: Now only addition and subtraction remain. Work from left to right: 45 + 9 = 54, then 54 − 30 = 24.
步骤3：此时只余下加法和减法。从左到右依次计算：45 + 9 = 54，然后 54 − 30 = 24。
The final answer is therefore 24. A very common error is to add 45 and 72 first, giving 117, and then continue incorrectly. Remember: division and multiplication always beat addition and subtraction unless brackets change the order.
最终答案是 24。一个非常常见的错误是先算 45 + 72 得到 117，然后再继续错误计算。请记住：除非有括号改变了顺序，否则除法和乘法总是优先于加法和减法。
Check: Another way to verify is to handle it as 45 + 9 + (−30) = 24. This reminds you that subtraction can be seen as adding a negative number, which helps avoid sign mistakes.
检验：另一种验证方法是将式子看作 45 + 9 + (−30) = 24。这提醒我们，减法可以看作是加上一个负数，有助于避免符号错误。
3. Question 2: Adding Fractions | 问题2：分数加法
Calculate 3/4 + 1/6 and give your answer in its simplest form.
Adding fractions with different denominators requires us to find a common denominator – a number that both 4 and 6 divide into exactly. The lowest common multiple of 4 and 6 is 12, so we will convert both fractions to twelfths.
分母不同的分数相加，需要找到一个公分母——一个能被 4 和 6 同时整除的数。4 和 6 的最小公倍数是 12，所以我们将两个分数都转化为以 12 为分母的分数。
Step 1: Convert 3/4. Multiply the numerator and denominator by 3: (3 × 3) / (4 × 3) = 9/12.
步骤1：转化 3/4。分子和分母同时乘以 3：(3 × 3)/(4 × 3) = 9/12。
Step 2: Convert 1/6. Multiply numerator and denominator by 2: (1 × 2) / (6 × 2) = 2/12.
步骤2：转化 1/6。分子和分母同时乘以 2：(1 × 2)/(6 × 2) = 2/12。
Step 3: Now add the fractions with the same denominator: 9/12 + 2/12 = 11/12. Keep the denominator the same and add the numerators.
步骤3：现在将同分母的分数相加：9/12 + 2/12 = 11/12。分母不变，分子相加。
The result is 11/12. This fraction is already in its simplest form because 11 and 12 have no common factors other than 1. A typical mistake is to add both numerators and denominators directly (3+1)/(4+6) = 4/10 = 2/5, which is completely wrong; denominators must be made equal first.
结果为 11/12。这个分数已经是最简形式，因为 11 和 12 除了 1 以外没有其他公因数。常见的错误是直接将分子和分母分别相加，(3+1)/(4+6) = 4/10 = 2/5，这是完全错误的方法；必须先让分母相等。
4. Question 3: Decimals and Percentages | 问题3：小数与百分数
(a) Write 0.35 as a percentage. (b) Find 25% of 80.
Percent means ‘per hundred’, so converting a decimal to a percentage involves multiplying by 100 and adding the % sign. Finding a percentage of an amount can be done by converting the percentage to a decimal or fraction and then multiplying.
百分数意为“每百”，因此将小数转化为百分数需要乘以 100 并加上百分号。求一个数的百分之几，可以先将百分数转化为小数或分数，然后再相乘。
Part (a): To write 0.35 as a percentage, multiply by 100: 0.35 × 100 = 35, so 0.35 = 35%. Moving the decimal point two places to the right gives the same result quickly.
(a) 部分：将 0.35 写成百分数，乘以 100：0.35 × 100 = 35，因此 0.35 = 35%。快速做法是将小数点向右移动两位。
Part (b): 25% means 25 out of 100, or the fraction 25/100 = 1/4, or the decimal 0.25. To find 25% of 80, multiply 0.25 × 80 = 20, or simply note that one quarter of 80 is 20.
(b) 部分：25% 表示每 100 份中的 25 份，即分数 25/100 = 1/4，或小数 0.25。求 80 的 25%，计算 0.25 × 80 = 20，或者直接想到 80 的四分之一是 20。
Both answers are linked: 0.35 = 35% and 25% of 80 = 20. Being comfortable switching between decimals, fractions and percentages is extremely useful for real‑life problems and further maths.
两个答案是关联的：0.35 = 35%，80 的 25% 是 20。能够熟练地在分数、小数和百分数之间切换，对解决实际问题以及后续数学学习非常有用。
5. Question 4: Algebra – Simplifying and Solving | 问题4：代数——化简与解方程
(a) Simplify 3a + 5b − a + 2b. (b) Solve 4x − 7 = 13.
In part (a), we collect like terms. ‘Like terms’ have exactly the same variable parts; here, 3a and −a are like terms, and 5b and 2b are like terms. Coefficients can be added while keeping the variable unchanged.
在 (a) 部分，我们合并同类项。“同类项”指含有完全相同字母部分的项；这里的 3a 和 −a 是同类项，5b 和 2b 是同类项。系数可以相加，字母部分保持不变。
Simplify: 3a − a = 2a, and 5b + 2b = 7b. So the simplified expression is 2a + 7b. Remember to keep the sign of each term when you rearrange.
化简：3a − a = 2a，5b + 2b = 7b。所以化简后的表达式为 2a + 7b。重组时要注意每一项的符号。
Part (b) is a one‑step equation. The goal is to isolate x. Start by undoing the subtraction of 7: add 7 to
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